#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Software: PyCharm
# @Version : Python-
# @Author  : Shengji He
# @Email   : hsjbit@163.com
# @File    : 4Sum2.py
# @Time    : 2020/11/27 10:04
# @Description:
from typing import List
from collections import Counter
from itertools import product


class Solution:
    def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
        """
        Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that
        A[i] + B[j] + C[k] + D[l] is zero.

        To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the
        range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

        Example:

            Input:
                A = [ 1, 2]
                B = [-2,-1]
                C = [-1, 2]
                D = [ 0, 2]

            Output:
                2

            Explanation:
                The two tuples are:
                1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
                2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

        :param A:
        :param B:
        :param C:
        :param D:
        :return:
        """
        countAB = Counter(u + v for u in A for v in B)
        ans = 0
        for u in C:
            for v in D:
                if -u - v in countAB:
                    ans += countAB[-u - v]
        return ans

    def fourSumCount2(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
        countAB = Counter(sum(n) for n in product(A, B))
        return sum(countAB.get(-sum(n), 0) for n in product(C, D))

    def fourSumCount3(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
        countAB = Counter(u + v for u in A for v in B)
        return sum(countAB.get(-u - v, 0) for u in C for v in D)


if __name__ == '__main__':
    A = [1, 2]
    B = [-2, -1]
    C = [-1, 2]
    D = [0, 2]
    S = Solution()
    print(S.fourSumCount(A, B, C, D))
    print(S.fourSumCount2(A, B, C, D))
    print(S.fourSumCount3(A, B, C, D))
    print('done')
